The Stacks project

53.12 Riemann-Hurwitz

Let $k$ be a field. Let $f : X \to Y$ be a morphism of smooth curves over $k$. Then we obtain a canonical exact sequence

\[ f^*\Omega _{Y/k} \xrightarrow {\text{d}f} \Omega _{X/k} \longrightarrow \Omega _{X/Y} \longrightarrow 0 \]

by Morphisms, Lemma 29.32.9. Since $X$ and $Y$ are smooth, the sheaves $\Omega _{X/k}$ and $\Omega _{Y/k}$ are invertible modules, see Morphisms, Lemma 29.34.12. Assume the first map is nonzero, i.e., assume $f$ is generically étale, see Lemma 53.12.1. Let $R \subset X$ be the closed subscheme cut out by the different $\mathfrak {D}_ f$ of $f$. By Discriminants, Lemma 49.12.6 this is the same as the vanishing locus of $\text{d}f$, it is an effective Cartier divisor, and we get

\[ f^*\Omega _{Y/k} \otimes _{\mathcal{O}_ X} \mathcal{O}_ X(R) = \Omega _{X/k} \]

In particular, if $X$, $Y$ are projective with $k = H^0(Y, \mathcal{O}_ Y) = H^0(X, \mathcal{O}_ X)$ and $X$, $Y$ have genus $g_ X$, $g_ Y$, then we get the Riemann-Hurwitz formula

\begin{align*} 2g_ X - 2 & = \deg (\Omega _{X/k}) \\ & = \deg (f^*\Omega _{Y/k} \otimes _{\mathcal{O}_ X} \mathcal{O}_ X(R)) \\ & = \deg (f) \deg (\Omega _{Y/k}) + \deg (R) \\ & = \deg (f) (2g_ Y - 2) + \deg (R) \end{align*}

The first and last equality by Lemma 53.8.4. The second equality by the isomorphism of invertible sheaves given above. The third equality by additivity of degrees (Varieties, Lemma 33.44.7), the formula for the degree of a pullback (Varieties, Lemma 33.44.11), and finally the formula for the degree of $\mathcal{O}_ X(R)$ (Varieties, Lemma 33.44.9).

To use the Riemann-Hurwitz formula we need to compute $\deg (R) = \dim _ k \Gamma (R, \mathcal{O}_ R)$. By the structure of zero dimensional schemes over $k$ (see for example Varieties, Lemma 33.20.2), we see that $R$ is a finite disjoint union of spectra of Artinian local rings $R = \coprod _{x \in R} \mathop{\mathrm{Spec}}(\mathcal{O}_{R, x})$ with each $\mathcal{O}_{R, x}$ of finite dimension over $k$. Thus

\[ \deg (R) = \sum \nolimits _{x \in R} \dim _ k \mathcal{O}_{R, x} = \sum \nolimits _{x \in R} d_ x [\kappa (x) : k] \]

with

\[ d_ x = \text{length}_{\mathcal{O}_{R, x}} \mathcal{O}_{R, x} = \text{length}_{\mathcal{O}_{X, x}} \mathcal{O}_{R, x} \]

the multiplicity of $x$ in $R$ (see Algebra, Lemma 10.52.12). Let $x \in X$ be a closed point with image $y \in Y$. Looking at stalks we obtain an exact sequence

\[ \Omega _{Y/k, y} \to \Omega _{X/k, x} \to \Omega _{X/Y, x} \to 0 \]

Choosing local generators $\eta _ x$ and $\eta _ y$ of the (free rank $1$) modules $\Omega _{X/k, x}$ and $\Omega _{Y/k, y}$ we see that $ \eta _ y \mapsto h \eta _ x $ for some nonzero $h \in \mathcal{O}_{X, x}$. By the exact sequence we see that $\Omega _{X/Y, x} \cong \mathcal{O}_{X, x}/h\mathcal{O}_{X, x}$ as $\mathcal{O}_{X, x}$-modules. Since the divisor $R$ is cut out by $h$ (see above) we have $\mathcal{O}_{R, x} = \mathcal{O}_{X, x}/h\mathcal{O}_{X, x}$. Thus we find the following equalities

\begin{align*} d_ x & = \text{length}_{\mathcal{O}_{X, x}}(\mathcal{O}_{R, x}) \\ & = \text{length}_{\mathcal{O}_{X, x}}(\mathcal{O}_{X, x}/h\mathcal{O}_{X, x}) \\ & = \text{length}_{\mathcal{O}_{X, x}}(\Omega _{X/Y, x}) \\ & = \text{ord}_{\mathcal{O}_{X, x}}(h) \\ & = \text{ord}_{\mathcal{O}_{X, x}}(“\eta _ y/\eta _ x") \end{align*}

The first equality by our definition of $d_ x$. The second and third we saw above. The fourth equality is the definition of $\text{ord}$, see Algebra, Definition 10.121.2. Note that since $\mathcal{O}_{X, x}$ is a discrete valuation ring, the integer $\text{ord}_{\mathcal{O}_{X, x}}(h)$ just the valuation of $h$. The fifth equality is a mnemonic.

Here is a case where one can “calculate” the multiplicity $d_ x$ in terms of other invariants. Namely, if $\kappa (x)$ is separable over $k$, then we may choose $\eta _ x = \text{d}s$ and $\eta _ y = \text{d}t$ where $s$ and $t$ are uniformizers in $\mathcal{O}_{X, x}$ and $\mathcal{O}_{Y, y}$ (Lemma 53.12.3). Then $t \mapsto u s^{e_ x}$ for some unit $u \in \mathcal{O}_{X, x}$ where $e_ x$ is the ramification index of the extension $\mathcal{O}_{Y, y} \subset \mathcal{O}_{X, x}$. Hence we get

\[ \eta _ y = \text{d}t = \text{d}(u s^{e_ x}) = e s^{e_ x - 1} u \text{d}s + s^{e_ x} \text{d}u \]

Writing $\text{d}u = w \text{d}s$ for some $w \in \mathcal{O}_{X, x}$ we see that

\[ ``\eta _ y/\eta _ x" = e s^{e_ x - 1} u + s^{e_ x} w = (e_ x u + s w)s^{e_ x - 1} \]

We conclude that the order of vanishing of this is $e_ x - 1$ unless the characteristic of $\kappa (x)$ is $p > 0$ and $p$ divides $e_ x$ in which case the order of vanishing is $> e_ x - 1$.

Combining all of the above we find that if $k$ has characteristic zero, then

\[ 2g_ X - 2 = (2g_ Y - 2)\deg (f) + \sum \nolimits _{x \in X} (e_ x - 1)[\kappa (x) : k] \]

where $e_ x$ is the ramification index of $\mathcal{O}_{X, x}$ over $\mathcal{O}_{Y, f(x)}$. This precise formula will hold if and only if all the ramification is tame, i.e., when the residue field extensions $\kappa (x)/\kappa (y)$ are separable and $e_ x$ is prime to the characteristic of $k$, although the arguments above are insufficient to prove this. We refer the reader to Lemma 53.12.4 and its proof.

slogan

Lemma 53.12.1. Let $k$ be a field. Let $f : X \to Y$ be a morphism of smooth curves over $k$. The following are equivalent

  1. $\text{d}f : f^*\Omega _{Y/k} \to \Omega _{X/k}$ is nonzero,

  2. $\Omega _{X/Y}$ is supported on a proper closed subset of $X$,

  3. there exists a nonempty open $U \subset X$ such that $f|_ U : U \to Y$ is unramified,

  4. there exists a nonempty open $U \subset X$ such that $f|_ U : U \to Y$ is étale,

  5. the extension $k(X)/k(Y)$ of function fields is finite separable.

Proof. Since $X$ and $Y$ are smooth, the sheaves $\Omega _{X/k}$ and $\Omega _{Y/k}$ are invertible modules, see Morphisms, Lemma 29.34.12. Using the exact sequence

\[ f^*\Omega _{Y/k} \longrightarrow \Omega _{X/k} \longrightarrow \Omega _{X/Y} \longrightarrow 0 \]

of Morphisms, Lemma 29.32.9 we see that (1) and (2) are equivalent and equivalent to the condition that $f^*\Omega _{Y/k} \to \Omega _{X/k}$ is nonzero in the generic point. The equivalence of (2) and (3) follows from Morphisms, Lemma 29.35.2. The equivalence between (3) and (4) follows from Morphisms, Lemma 29.36.16 and the fact that flatness is automatic (Lemma 53.2.3). To see the equivalence of (5) and (4) use Algebra, Lemma 10.140.9. Some details omitted. $\square$

Lemma 53.12.2. Let $f : X \to Y$ be a morphism of smooth proper curves over a field $k$ which satisfies the equivalent conditions of Lemma 53.12.1. If $k = H^0(Y, \mathcal{O}_ Y) = H^0(X, \mathcal{O}_ X)$ and $X$ and $Y$ have genus $g_ X$ and $g_ Y$, then

\[ 2g_ X - 2 = (2g_ Y - 2) \deg (f) + \deg (R) \]

where $R \subset X$ is the effective Cartier divisor cut out by the different of $f$.

Lemma 53.12.3. Let $X \to \mathop{\mathrm{Spec}}(k)$ be smooth of relative dimension $1$ at a closed point $x \in X$. If $\kappa (x)$ is separable over $k$, then for any uniformizer $s$ in the discrete valuation ring $\mathcal{O}_{X, x}$ the element $\text{d}s$ freely generates $\Omega _{X/k, x}$ over $\mathcal{O}_{X, x}$.

Proof. The ring $\mathcal{O}_{X, x}$ is a discrete valuation ring by Algebra, Lemma 10.140.3. Since $x$ is closed $\kappa (x)$ is finite over $k$. Hence if $\kappa (x)/k$ is separable, then any uniformizer $s$ maps to a nonzero element of $\Omega _{X/k, x} \otimes _{\mathcal{O}_{X, x}} \kappa (x)$ by Algebra, Lemma 10.140.4. Since $\Omega _{X/k, x}$ is free of rank $1$ over $\mathcal{O}_{X, x}$ the result follows. $\square$

Lemma 53.12.4. Notation and assumptions as in Lemma 53.12.2. For a closed point $x \in X$ let $d_ x$ be the multiplicity of $x$ in $R$. Then

\[ 2g_ X - 2 = (2g_ Y - 2) \deg (f) + \sum \nolimits d_ x [\kappa (x) : k] \]

Moreover, we have the following results

  1. $d_ x = \text{length}_{\mathcal{O}_{X, x}}(\Omega _{X/Y, x})$,

  2. $d_ x \geq e_ x - 1$ where $e_ x$ is the ramification index of $\mathcal{O}_{X, x}$ over $\mathcal{O}_{Y, y}$,

  3. $d_ x = e_ x - 1$ if and only if $\mathcal{O}_{X, x}$ is tamely ramified over $\mathcal{O}_{Y, y}$.

Proof. By Lemma 53.12.2 and the discussion above (which used Varieties, Lemma 33.20.2 and Algebra, Lemma 10.52.12) it suffices to prove the results on the multiplicity $d_ x$ of $x$ in $R$. Part (1) was proved in the discussion above. In the discussion above we proved (2) and (3) only in the case where $\kappa (x)$ is separable over $k$. In the rest of the proof we give a uniform treatment of (2) and (3) using material on differents of quasi-finite Gorenstein morphisms.

First, observe that $f$ is a quasi-finite Gorenstein morphism. This is true for example because $f$ is a flat quasi-finite morphism and $X$ is Gorenstein (see Duality for Schemes, Lemma 48.25.7) or because it was shown in the proof of Discriminants, Lemma 49.12.6 (which we used above). Thus $\omega _{X/Y}$ is invertible by Discriminants, Lemma 49.16.1 and the same remains true after replacing $X$ by opens and after performing a base change by some $Y' \to Y$. We will use this below without further mention.

Choose affine opens $U \subset X$ and $V \subset Y$ such that $x \in U$, $y \in V$, $f(U) \subset V$, and $x$ is the only point of $U$ lying over $y$. Write $U = \mathop{\mathrm{Spec}}(A)$ and $V = \mathop{\mathrm{Spec}}(B)$. Then $R \cap U$ is the different of $f|_ U : U \to V$. By Discriminants, Lemma 49.9.4 formation of the different commutes with arbitrary base change in our case. By our choice of $U$ and $V$ we have

\[ A \otimes _ B \kappa (y) = \mathcal{O}_{X, x} \otimes _{\mathcal{O}_{Y, y}} \kappa (y) = \mathcal{O}_{X, x}/(s^{e_ x}) \]

where $e_ x$ is the ramification index as in the statement of the lemma. Let $C = \mathcal{O}_{X, x}/(s^{e_ x})$ viewed as a finite algebra over $\kappa (y)$. Let $\mathfrak {D}_{C/\kappa (y)}$ be the different of $C$ over $\kappa (y)$ in the sense of Discriminants, Definition 49.9.1. It suffices to show: $\mathfrak {D}_{C/\kappa (y)}$ is nonzero if and only if the extension $\mathcal{O}_{Y, y} \subset \mathcal{O}_{X, x}$ is tamely ramified and in the tamely ramified case $\mathfrak {D}_{C/\kappa (y)}$ is equal to the ideal generated by $s^{e_ x - 1}$ in $C$. Recall that tame ramification means exactly that $\kappa (x)/\kappa (y)$ is separable and that the characteristic of $\kappa (y)$ does not divide $e_ x$. On the other hand, the different of $C/\kappa (y)$ is nonzero if and only if $\tau _{C/\kappa (y)} \in \omega _{C/\kappa (y)}$ is nonzero. Namely, since $\omega _{C/\kappa (y)}$ is an invertible $C$-module (as the base change of $\omega _{A/B}$) it is free of rank $1$, say with generator $\lambda $. Write $\tau _{C/\kappa (y)} = h\lambda $ for some $h \in C$. Then $\mathfrak {D}_{C/\kappa (y)} = (h) \subset C$ whence the claim. By Discriminants, Lemma 49.4.8 we have $\tau _{C/\kappa (y)} \not= 0$ if and only if $\kappa (x)/\kappa (y)$ is separable and $e_ x$ is prime to the characteristic. Finally, even if $\tau _{C/\kappa (y)}$ is nonzero, then it is still the case that $s \tau _{C/\kappa (y)} = 0$ because $s\tau _{C/\kappa (y)} : C \to \kappa (y)$ sends $c$ to the trace of the nilpotent operator $sc$ which is zero. Hence $sh = 0$, hence $h \in (s^{e_ x - 1})$ which proves that $\mathfrak {D}_{C/\kappa (y)} \subset (s^{e_ x - 1})$ always. Since $(s^{e_ x - 1}) \subset C$ is the smallest nonzero ideal, we have proved the final assertion. $\square$

Comments (2)

Comment #5012 by Elyes Boughattas on

Typo in the Riemann-Hurwitz formula preceding lemma 0C1C: should be .

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